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I am interested in the limit as $x\rightarrow\infty$ of the following function:


Here $0<a<1$, $0<b\ll 1$ and $0<c\ll 1$ are small positive constants, and $d$ is a positive variable.

I did some numerical evaluations of this expression, and it seems to me that for $0<d<1/2$, $\lim_{x\rightarrow\infty}f(x)=0$; for $d=1/2$, $\lim_{x\rightarrow\infty}f(x)=e^{g(a,b,c)}$; and for $d>1/2$, $\lim_{x\rightarrow\infty}f(x)=1$. But I am stuck trying to prove that.

I tried putting the expression into the following form:


and then applying the usual method of taking its log to obtain the following:

$$\begin{array}{rcl}\log f(x)&=&-\frac{x}{2}\left(\log(1+ab)+\log(1+ab+(1-a)cx^{-d})\right)\\ & &-x\log\left(1-\sqrt{\frac{ab(ab+(1-a)cx^{-d})}{(1+ab)(1+ab+(1-a)cx^{-d})}}\right)\\ &\approx&-\frac{x}{2}\left(2ab+(1-a)cx^{-d}\right)+x\sqrt{\frac{ab(ab+(1-a)cx^{-d})}{(1+ab)(1+ab+(1-a)cx^{-d})}}\end{array}$$

but I don't know what to do with the square root...

Any hints would be appreciated.

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2 Answers 2

up vote 1 down vote accepted

I solved this, and figured that I should post the answer here.

Taking $\log f(x)$ (in the original form) yields:

$$\log f(x)=-x\log\left(\sqrt{(1+ab)(1+ab+(1-a)cx^{-d})}-\sqrt{ab(ab+(1-a)cx^{-d})}\right)$$

Now, $f(x)=\exp[\log f(x)]$, but we can take the limit through the exponent since it's continuous.

Substituting $y=1/x$ and taking the limit as $y\rightarrow 0$ results in an indeterminate form 0/0. However, that allows us to use L'Hopital's rule in the above expression. For the sake of clarity, let's also substitute $u=ab$ and $v=(1-a)c$. The new expression looks as follows:

$$\lim_{y\rightarrow0}\log f(y)=-\lim_{y\rightarrow0}\frac{\log\left(\sqrt{(1+u)(1+u+vy^{d})}-\sqrt{u(u+vy^{d})}\right)}{y}$$

Per L'Hopital's rule, we take the derivative with respect to $y$ of the numerator and denominator and evaluate the limit of their ratio. Since the derivative of the denominator is 1, we are interested in the limit of the derivative of the numerator. The expression for it is as follows:

$$\lim_{y\rightarrow0}\frac{\frac{d (1+u) v y^{d-1}}{2 \sqrt{(1+u) \left(1+u+v y^d\right)}}-\frac{d u v y^{d-1}}{2 \sqrt{u^2+u v y^d}}}{\sqrt{(1+u) \left(1+u+v y^d\right)}-\sqrt{u^2+u v y^d}}$$

The above looks daunting but it can be simplified by multiplying the numerator and the denominator of the large fraction by $\frac{d (1+u) v y^{d-1}}{2 \sqrt{(1+u) \left(1+u+v y^d\right)}}+\frac{d u v y^{d-1}}{2 \sqrt{u^2+u v y^d}}$. Doing some tedious algebra results in the following expression:

$$L=\lim_{y\rightarrow0}\frac{d v^2 y^{2 d-1}}{2 \left(u+u^2+2 u v y^d+v y^d \left(1+v y^d\right)+\sqrt{u \left(u+v y^d\right)} \sqrt{(1+u) \left(1+u+v y^d\right)}\right)}$$.

Now, $\lim f(x)=\exp(-L)$. Thus, we can analyze the limit as $y\rightarrow0$ (or, correspondingly, $x\rightarrow\infty$) that we originally set out to analyze by looking at $L$ for different cases of $d$. When $d<1/2$, $L\rightarrow\infty$ and $\lim f(x)=0$; when $d>1/2$, $L=0$ and $\lim f(x)=1$; when $d=1/2$, $L=\frac{v^2}{8(u+u^2)}$ and $\lim f(x)=\exp\left(-\frac{v^2}{8(u+u^2)}\right)$, as my numerical experiments suggested.

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Square root is continuous. You can leave it as it is.

If $x\to\infty$, then with the given assumptions, $x^{-d}\to 0$, $\ (1-a),c\ne 0$, and hence $$\sqrt{\frac{ab(ab+(1-a)cx^{-d})}{(1+ab)(1+ab+(1-a)cx^{-d})}} \ \to \sqrt{\frac{(ab)^2}{(1+ab)^2}}$$

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